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Chapter 9: Problem 3

The distribution of the scores on a certain exam is \(N(80,5)\) which means that the exam scores are Normally distributed with a mean of 80 and a standard deviation of 5 . a. Sketch or use technology to create the curve and label on the \(x\) -axis the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be greater than 90\. Shade the region under the Normal curve whose area corresponds to this probability.

Short Answer

Expert verified
The probability that a randomly selected score will be greater than 90 is roughly 0.0228 or 2.28%.

Step by step solution

01

Understanding Normal Distribution

A normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In this case, the mean (\(\mu\)) is 80 and the standard deviation (\(\sigma\)) is 5. These represent key points on the distribution graph.
02

Sketch the Normal Distribution Curve

Draw a curve on the x-axis. At the center of the curve is the mean (80). Then label the following points on the x-axis from left to right: \(\mu - 3\sigma,(80 - 3*5) = 65\); \(\mu - 2\sigma,(80 - 2*5) = 70\); \(\mu - \sigma,(80 - 5) = 75\); \(\mu,(80)\); \(\mu + \sigma,(80 + 5) = 85\); \(\mu + 2\sigma,(80 + 2*5) = 90\); \(\mu + 3\sigma,(80 + 3*5) = 95\)
03

Calculate the z-score

The z-score for a number \(x\) in a Normal distribution can be calculated with the following formula: \(Z = \frac{(X - \mu)}{\sigma}\). Plug in the score 90 to calculate the z-score which is \(Z = \frac{(90-80)}{5} = 2\).
04

Find the Probability That a Score Will Be Greater Than 90

We can find the area to the left of Z using Z-tables, which provides the probability that a value selected at random is less than Z. Looking up a Z of 2 in the Z-table gives us approximately 0.9772. As we need to find the probability of a score being greater than 90, that would mean finding the area to the right of Z. It can be calculated by taking 1 minus the probability of the score being less than 90 (which was found using the z-table). So, \(P(X>90) = 1- P(X<90) = 1-0.9772 = 0.0228\). This indicates that there is a 2.28% chance of a randomly selected score being greater than 90.
05

Shade the area under the curve

On your graph, shade the region to the right of the score 90. This shaded area represents the probability that a randomly selected score is greater than 90.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a fundamental concept in statistics and mathematics that deals with the likelihood or chance of an event occurring. It ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty. When we talk about probability in a normal distribution, we refer to the area under the curve that signifies how likely the scores are to fall within a certain range.
  • In a normal distribution, such as the one with a mean of 80 and a standard deviation of 5 given in the exercise, most of the data points cluster around the mean.
  • The total area under the curve is equal to 1, representing the certainty that the score will land somewhere on the x-axis.
  • When calculating probabilities in normal distribution, we often focus on certain ranges or thresholds, like finding the likelihood of a score being greater than a particular number.
In the exercise, finding the probability that a score is greater than 90 involves determining the area under the curve to the right of 90. With the mean at 80 and using the z-score to quantify the distance of 90 from the mean, we use probability tables to find these areas.
Z-score
A Z-score is a measure used in statistics to determine the number of standard deviations a data point is from the mean of a set of data. It is a way to standardize scores on a normal distribution, making it easier to compare different data points.
  • The formula to calculate a Z-score is given by:
    \[ Z = \frac{(X - \mu)}{\sigma} \]
    where \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • A Z-score tells you how far from the mean your data point is: a Z-score of 0 indicates that the data point is exactly at the mean, while positive or negative Z-scores indicate positions on either side of the mean.
In the context of the exercise, 90 was the score we were interested in. Using the formula, the Z-score calculated was 2, meaning 90 is 2 standard deviations above the mean of 80. This helps in finding the probability related to 90 by looking up Z-score tables.
Standard Deviation
Standard deviation is a measure of the spread or dispersion of a set of values. In a normal distribution, it gives us insight into how much the individual data points deviate from the mean, in essence quantifying variability.
  • In cases like the given normal distribution with a standard deviation of 5, it means that the scores typically fall within 5 units of the mean.
  • Standard deviation is a crucial parameter in normal distributions, as it plays a key role in forming the shape of the bell curve: a smaller standard deviation results in a sharper peak, while a larger value indicates a flatter spread.
  • It helps identify how much variation or "spread out" the values are from the average value (the mean).
Understanding the standard deviation allows us to predict the likelihood of various outcomes. When a score like 90 is two standard deviations away (as calculated in the previous section), it stands outside the typical range, indicating a smaller probability of occurrence in our distribution.

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Most popular questions from this chapter

In exercise \(9.31\), two intervals were given for the same data, one for \(95 \%\) confidence and one for \(90 \%\) confidence. a. How would a \(99 \%\) interval compare? Would it be narrower than both. wider than both, or between the two in width. Explain. b. If we wanted to use a \(99 \%\) confidence level and get a narrower width. how could we change our data collection?

State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a \(95 \%\) level of confidence to a \(90 \%\) level of confidence b. Changing from a sample size of 30 to a sample size of 20 c. Changing from a standard deviation of 3 inches to a standard deviation of \(2.5\) inches

According to home-water-works.org, the average shower in the United States lasts \(8.2\) minutes. Assume this is correct, and assume the standard deviation of 2 minutes. a. Do you expect the shape of the distribution of shower lengths to be Normal, right-skewed, or left-skewed? Explain. b. Suppose that we survey a random sample of 100 people to find the length of their last shower. We calculate the mean length from the sample and record the value. We repeat this 500 times. What will be the shape of the distribution of these sample means? c. Refer to part b. What will be the mean and the standard deviation of the distribution of these sample means?

Drivers in Wyoming drive more miles yearly than motorists in any other state. The annual number of miles driven per licensed driver in Wyoming is 22,306 miles. Assume the standard deviation is 5500 miles. A random sample of 200 licensed drivers in Wyoming is selected and the mean number of miles driven yearly for the sample is calculated. (Source: 2017 World Almanac and Book of Facts) a. What value would we expect for the sample mean? b. What is the standard error for the sample mean?

A random sample of 10 independent healthy people showed the following body temperatures (in degrees Fahrenheit): $$98.5,98.299 .0,96.3,98.3,98.7,97.2,99.1,98.7,97.2$$ Test the hypothesis that the population mean is not \(98.6^{\circ} \mathrm{F}\), using a significance level of \(0.05 .\) See page 500 for guidance.
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